Title: InkPhysics / Nuclear physics
On every note of this deck, unless a note says otherwise:
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Front: DEFINITION
Define: Decay constant.
Back: The probability per unit time that a given nucleus decays: λ, in s^−1.

Front: DEFINITION
Define: Induced fission.
Back: A heavy nucleus splitting into two smaller nuclei after absorbing a neutron, releasing energy and further neutrons.

Front: EQUATION — NOT ON THE DATA SHEET
Activity falling with time:
A = ?
Back: A = A_0e^-λt

Front: DEFINITION
Define: Activity.
Back: The number of nuclei in a sample decaying per second: A = λN, in becquerels.

Front: CHECK YOURSELF
The nuclear mass of iron-56 is 55.92066 u. Using m_p = 1.00728 u and m_n = 1.00867 u, find the mass difference, the binding energy, and the binding energy per nucleon.
Show a hint
Iron-56 has 26 protons and 30 neutrons; parts minus whole first.
Back: Separate parts: 26 × 1.00728 + 30 × 1.00867 = 56.44938 u, so Δm = 56.44938 − 55.92066 = 0.52872 u.
Binding energy = 0.52872 × 931.5 = 492 MeV.
Per nucleon: 492 / 56 = 8.8 MeV: the very top of the curve, which is why iron is where both fusion and fission run out of road.

Front: EQUATION — NOT ON THE DATA SHEET
Neutron energy kept in an elastic moderator collision:
fraction kept = ?
Back: fraction kept = (M - mM + m)^2

Front: EQUATION — NOT ON THE DATA SHEET
Nuclear density from R₀:
ρ = ?
Back: ρ = 3u4πR_0^3

Front: DEFINITION
Define: Mass defect.
Back: The difference between the mass of a nucleus and the total mass of its separate nucleons.

Front: CHECK YOURSELF
Caesium-137 has a half-life of 30 years. A sealed source contains 3.0 μg of caesium-137 (molar mass 137 g mol⁻¹). Find the decay constant and the activity of the source. Take 1 year = 3.15 × 10⁷ s.
Show a hint
Half-life to λ first; then mass to moles to nuclei; then A = λN.
Back: T½ = 30 × 3.15 × 10^7 = 9.45 × 10^8 s, so λ = ln 2 / (9.45 × 10^8) = 7.3 × 10^-10 s^-1.
N = (3.0 × 10^-6 / 137) × 6.02 × 10^23 = 1.3 × 10^16 nuclei.
A = λN = 7.3 × 10^-10 × 1.3 × 10^16 = 9.7 × 10^6 Bq: ten million decays a second from three millionths of a gram.

Front: CHECK YOURSELF
Take R₀ = 1.2 fm. Find the radius of an iron-56 nucleus, and then its density, given the atomic mass unit u = 1.661 × 10⁻²⁷ kg. Comment on how the density would differ for gold-197.
Show a hint
Radius first from the cube-root law; then mass over the volume of a sphere.
Back: R = R_0A^1/3 = 1.2 × 56^1/3 = 1.2 × 3.83 = 4.6 fm.
Mass = 56u = 9.3 × 10^-26 kg; volume = (4/3)π(4.59 × 10^-15)^3 = 4.1 × 10^-43 m^3.
ρ = 9.3 × 10^-26 / 4.1 × 10^-43 = 2.3 × 10^17 kg m^-3, and gold's comes out the same: A cancels, so every nucleus shares this density.

Front: DEFINITION
Define: Half-life.
Back: The time taken for the number of undecayed nuclei in a sample, and so its activity, to halve.

Front: CHECK YOURSELF
A gamma source gives a corrected count rate of 1440 counts per minute at 0.50 m. Predict the corrected count rate at 1.50 m, and explain why the background was subtracted before either reading was used.
Show a hint
How many times further than 1.50 m is the new distance, and what does the square of that factor do?
Back: The distance is 3 times greater, so the intensity falls by a factor of 3^2 = 9.
1440 / 9 = 160 counts per minute.
The inverse-square law applies to the source's own photons; background counts come from everywhere and do not fall with x, so they must be removed first.

Front: DEFINITION
Define: Moderator.
Back: The reactor material that slows fission neutrons by elastic collisions to thermal speeds, at which they are more likely to cause further fission.

Front: CHECK YOURSELF
Deuterium (2.01355 u) and tritium (3.01550 u) fuse to form helium-4 (4.00150 u) and a neutron (1.00867 u), masses given as nuclear masses. Find the energy released.
Show a hint
Total mass before, total after, difference times 931.5.
Back: Before: 2.01355 + 3.01550 = 5.02905 u. After: 4.00150 + 1.00867 = 5.01017 u.
Δm = 5.02905 − 5.01017 = 0.01888 u.
E = 0.01888 × 931.5 = 17.6 MeV, carried off as kinetic energy of the helium nucleus and, mostly, the neutron.

Front: EQUATION — NOT ON THE DATA SHEET
Closest approach: the alpha's kinetic energy:
E_k = ?
Back: E_k = Qq4πε_0r_min

Front: DEFINITION
Define: Binding energy.
Back: The energy needed to separate a nucleus into its constituent nucleons, equal to the mass defect times c^2.

Front: DEFINITION
Define: Critical mass.
Back: The minimum mass of fissile material in which, on average, one neutron per fission goes on to cause another fission.

Front: CHECK YOURSELF
In a head-on elastic collision with a stationary nucleus of mass M, a neutron of mass m keeps the fraction ((M − m)/(M + m))² of its kinetic energy. Compare a collision with hydrogen (M = m) and with carbon-12 (M = 12m), and hence explain why water moderates in fewer collisions than graphite.
Show a hint
Substitute the two masses; then think about what the fractions mean per bounce.
Back: Hydrogen: (0/2m)^2 = 0: a head-on strike hands over every joule in one collision.
Carbon-12: (11m/13m)^2 = 121/169 = 0.72: the neutron keeps 72%, losing only about a quarter per head-on bounce.
Real collisions are rarely head-on, so several are always needed, but hydrogen-rich water still thermalises neutrons in far fewer collisions than graphite: the lighter the target, the bigger each bite.